Hello, hachiko!

Another approach . . .

Equilateral triangle $\displaystyle ABC$ is inscribed in a circle.

The angle bisector at $\displaystyle A$ intersects the circle again at $\displaystyle P$ ($\displaystyle AP$ is adiameter).

If the length of segment $\displaystyle AC$ is 10, find: .$\displaystyle AP - (BP + CP)$ Code:

A
* o *
* /|\ *
* / | \ *
* / | \ 10 *
/ |30°\
* / | \ *
* / | \ *
* / | \ *
/ | \
B o- - - - + - - - -o C
* | * *
* 60° | * *
* o *
P

We have: inscribed equilateral triangle $\displaystyle ABC\!:\;\;AC = 10$

. . with diameter $\displaystyle AP,\;\;\angle CAP = 30^o.$

Draw chord $\displaystyle CP$.

$\displaystyle \Delta ACP$ is inscribed in a semicircle.

. . Hence: .$\displaystyle \angle ACP = 90^o,\;\;\angle APC = 60^o$

We have: .$\displaystyle CP = x,\;\;AC = 10,\;\;AP = 2x$ .(30°-60° right triangle)

Pythagorus says: .$\displaystyle CP^2 + AC^2 \:=\:AP^2 \quad\Rightarrow\quad x^2 + 10^2 \:=\:(2x)^2 $

. . $\displaystyle x^2 + 100 \:=\:4x^2 \quad\Rightarrow\quad 3x^2 \:=\:100 \quad\Rightarrow\quad x^2 \:=\:\frac{100}{3}\quad\Rightarrow\quad x \:=\:\frac{10\sqrt{3}}{3}$

Hence: .$\displaystyle BP \,=\,CP \,=\,\frac{10\sqrt{3}}{3},\;\;AP \,=\,\frac{20\sqrt{3}}{3}$

Therefore: .$\displaystyle AP - (BP + CP) \;\;=\;\;\frac{20\sqrt{3}}{3} - \left(\frac{10\sqrt{3}}{3} + \frac{10\sqrt{3}}{3}\right) \;\;=\;\;0 $